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Classical mechanics (find trajectory and kinetic energy)

  • Thread starter NanoMath
  • Start date
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1. Homework Statement
Given the force ## \vec{ F }(x) = (-12x + 6) \vec{i} ## ; find kinetic energy ## T## at the point ##x=2## and trajectory of a particle ## \vec{r}(t) ##, given that ## \vec{r}(t=0)=\vec{0}## and ##\dot{\vec{r}}(t=0)=\vec{0}## .

3. The Attempt at a Solution

Since ##\nabla\times \vec{F} = \vec{0}## we can find potential function ## \vec{F}(x) = -\nabla U(x)## , therefore ## U(x)= 6x²-6x + C ## . From here I am not sure how to proceed.
Second idea is to solve differential equation but I'm not sure how to do it because force is a function of ## x## and not ## t ## .
$$ m\ddot{x}=-12x+6$$ $$\ddot{x}+\frac{1}{m}12x-\frac{1}{m}6=0$$ let's suppose ##m=1## for simplicity, then we get $$ \ddot{x}+12x-6=0$$
 

haruspex

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1. Homework Statement
Given the force ## \vec{ F }(x) = (-12x + 6) \vec{i} ## ; find kinetic energy ## T## at the point ##x=2## and trajectory of a particle ## \vec{r}(t) ##, given that ## \vec{r}(t=0)=\vec{0}## and ##\dot{\vec{r}}(t=0)=\vec{0}## .

3. The Attempt at a Solution

Since ##\nabla\times \vec{F} = \vec{0}## we can find potential function ## \vec{F}(x) = -\nabla U(x)## , therefore ## U(x)= 6x²-6x + C ## . From here I am not sure how to proceed.
Second idea is to solve differential equation but I'm not sure how to do it because force is a function of ## x## and not ## t ## .
$$ m\ddot{x}=-12x+6$$ $$\ddot{x}+\frac{1}{m}12x-\frac{1}{m}6=0$$ let's suppose ##m=1## for simplicity, then we get $$ \ddot{x}+12x-6=0$$
With a simple change of variable you can eliminate the constant 6 from the differential equation and obtain a very well known form. Think springs.
By the way, this belongs in Introductory Physics homework.
 

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